Question

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surfaces are of the same radius of curvature $R\, = \,14\, cm$.For this bi-convex lens, for an object distance of $40\, cm$, the image distance will be

• A. $- 280.0 \,cm$
• B. $40.0\, cm$
• C. $21.5\, cm$
• D. $13.3\, cm$

Answer 1

Answer: (B)

Using the lens formula,
$\frac{1}{v}-\frac{1}{u}=\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}$
or $\frac{1}{v}=\frac{1}{u}+\frac{1}{f_1}+\frac{1}{f_2}$
$15\,mm =\frac{1}{u}+(n_1-1)\Bigg(\frac{1}{R_1}-\frac{1}{R_2}\Bigg)+(n_2-1)\Bigg(\frac{1}{R'_1}-\frac{1}{R'_2}\Bigg)$
Substituting the values, we get
$\frac{1}{v}=\frac{1}{-40}+(1.5-1)\Bigg(\frac{1}{14}-\frac{1}{\infty}\Bigg)+(1.2-1)\Bigg(\frac{1}{\infty}-\frac{1}{14}\Bigg)$
Solving this equations, we get
$v = + 40 \,cm$